Question

If $3^{rd}$, $7^{th}$ and $12^{th}$ terms of an $A.P.$ are three consecutive terms of a $G.P.$, then the common ratio of the $G.P.$ is

• A. $\frac{5}{4}$
• B. $\frac{9}{4}$
• C. $\frac{2}{9}$
• D. $\frac{1}{2}$
• E. $\frac{12}{7}$

Answer 1

Answer: (A)

Let first term and common difference of an AP are $a$ and $d$.
$\therefore T_3=a+2d$
$T_7=a+6d$
and $T_{12}=a+11d$
But $T_3,T_7$ and $T_{12}$ are consecutive terms of an GP.
$\therefore (a+6d{)}^2=(a+2d)(a+11d)$
$a^2+36d^2+12ad=a^2+13ad+22d^2$
$ad=14d^2$
$\Rightarrow a=14d$
$\therefore$ Terms are $14d+2d,14d+6d,14d+11d$
i.e., $16d,20d,25d$
$\therefore$ Common ratio $=\frac{20d}{16d}=\frac{5}{4}$